Answer
$(5,2)$
Work Step by Step
Step 1
Let us find the point which is $\frac{2}{5}$ away from the line segment formed by the points
$P_1(3,6)$ and
$P_2(8,-4)$.
Step 2
We let a point $P(x,y)$ be the point on $\overrightarrow{P_1P_2}$ that is $\frac{2}{5}$ away from the given points. Since $\overrightarrow{P_1P_2}$ is given by:
$\overrightarrow{P_1P_2}=\langle 8-3,-4-6\rangle=\langle 5,-10\rangle$
the parametric equations for the point $P(x,y)$ with $P_1(3,6)$
as the reference point is given by:
$\langle x,y\rangle = \langle x_0,y_0\rangle + t\langle 5,-10\rangle$
where $\langle x_0,y_0\rangle$ is $P_1$, namely $(3,6)$.
Thus,
$x=x_0+5t;\ y=y_0-10t$
Step 3
It is given that the line segment $\overline{PP_1}$ is $\frac{2}{5}$ away from $P_1$.
Thus,
$\overrightarrow{PP_1}=\frac{2}{5}(\overrightarrow{P_1P_2})$
By observation, $\overrightarrow{PP_1}$ and $\overrightarrow{P_1P_2}$ are parallel to each other. Then that means that our $t=\frac{5}{2}$.
Using the parametric equations we found,
$\langle x,y\rangle = \langle x_0,y_0\rangle + t\langle 5,-10\rangle = \langle 3,6\rangle + \frac{5}{2}\langle 5,-10\rangle = \langle 5,2\rangle$
Thus the point $P$ that is $\frac{2}{5}$ of the way from $P_1$
is $(5,2)$.
Result: $P(5,2)$
